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My question:

Define the distance on $C[0,1]$ by $$d(f,g)=\max_{x\in [0,1]}|f(x)-g(x)|.$$ Prove that the set $S \subset C[0,1]$ is not compact, where $S=\{f \in C[0,1]|d(f,0)=1\}$.

My idea: For this, I assume that if I choose a sequence $\{f_n\}$ in $S$, that is Cauchy, but the limit of $\{f_n\}$ is not continuous. This implies $S$ is not compact.

I chose $f_n(x)=(-1)^{n}$. The limit points of $f_n(x)$ are $f(x)=1,$ and $-1$. I observed that this sequence is not Cauchy, so I cannot use my idea.

I choose another example, $f_n(x)=x^n$, but this is also not Cauchy.

Can anyone suggest an example of this case that works?

Calvin Khor
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User124356
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    A sequence of continuous functions that is cauchy with respect to d will be continuous, since $(C[0,1],d)$ is a complete metric space. Maybe instead try to create a sequence with no convergent subsequence – Calvin Khor Jul 02 '20 at 03:30
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    Your last example can be used to prove the result in a different way: it’s a sequence that has no convergent subsequence with respect to the metric $d$. The accepted answer to this question proves the result in yet another way. – Brian M. Scott Jul 02 '20 at 03:36

1 Answers1

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Here is one indirect way:

Let $\phi(f) = \int_{0}^{1/2} f(t) dt-\int_{1/2}^{1} f(t) dt$. It is straightforward to see that $\phi$ is continuous on $C[0,1]$ and $ \phi(f) < 1$ for all $f \in S$.

Let $f_n$ be the function whose graph is given by joining the points $(0,{1}), ({1 \over 2}-{1 \over n},1), ({1 \over 2}+{1 \over n},-1), (1,-1)$ and note that $f_n \in S$ and $\phi(f_n) \to 1$.

Hence $\sup_{f \in S} \phi(f) = 1$ but the $\sup$ is not attained. If $S$ was compact then there would be a maximiser.

copper.hat
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